MinnieMinnie
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- Nov 4, 2018
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- HSC
- 2020
sin-1(0.95/1.91)View attachment 28412
I know how to do everything but (c)
The answer is supposed to be 0.52
(a) 1.91
(b) 0.95m
(d) 0.5
How would i work out c?
Ohhhh Ok thank you!!!!sin-1(0.95/1.91)
Hey where's my logic wrong here. One period of the graph is 720 degrees. So 0.52x2 = 1.04 for b?At , and you know that , so solve:
.
This needs to be done in radians. The value of b gives the shift of an x intercept away from the origin.Hey where's my logic wrong here. One period of the graph is 720 degrees. So 0.52x2 = 1.04 for b?
ah ok, that makes sense. But doesn't it look like its shifted by 104 degrees since it's shifted by 2/3 of the first quarter of a period.This needs to be done in radians. The value of b gives the shift of an x intercept away from the origin.
y=sin(nt+b)The zero point is at
which corresponds to your idea, but it doesn't alter the value of itself which was found when and so was not dependent on .
If I take the question as having the function: with and given that at we have then:y=sin(nt+b)
= sin(n(t+b/n))
At 0.5, b/n = 2b. Inside the brackets is t+2b, therefore the graph of t+2b changes by 2b meaning that if a graph where n =1 is sketched out, it changes by two b. Now it is a matter of applying the multiple n, which basically horizontally shrinks the graph by a factor of 2, thereby making the change only b. However, as it is horizontally shrunk, it appears as if the same proportion of the curve for one period of the curve has been moved.
MinnieMinnie, may I ask where this came from?View attachment 28412
I know how to do everything but (c)
The answer is supposed to be 0.52
(a) 1.91
(b) 0.95m
(d) 0.5
How would i work out c?
This was on our HW sheet!!!MinnieMinnie, may I ask where this came from?
Everyone should note that the way this curve is drawn, with apparent vertical inflexions at each -intercept, is not how a sine function should be drawn.
making the tangents at the intecepts diagonal.
In this specific case:
and so the -intercepts are close to having a gradient of 1... which is not at all how the diagram has been drawn by the source.
Thanks for replying, MinnieMinnie. Wherever your school / teacher got the question from, the illustration of the trig function is awful.This was on our HW sheet!!!